Wire diameter reduction

With a single pull, the wire diameter is reduced by 10%. How many draws are required to minimize the wire diameter from 5 mm to less than half?

Final Answer:

n =  7

Step-by-step explanation:

$$\begin{gathered} q=1/2=\frac{1}{2}=0.5 \\ {q}_1=1-10/100=\frac{9}{10}=0.9 \\ {q}_1^n=q \\ n\log q_1=\log q \\ {n}_1=\frac{\log q}{\log (q_1)}=\frac{\log 0.5}{\log 0.9}\doteq 6.5788 \\ n=\lceil n_1\rceil =\lceil 6.5788\rceil =7 \end{gathered}$$

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